Answer
Verified
412.2k+ views
Hint: To find the radius of the incircle, first find the area of the triangle using the formula, $\dfrac{1}{2}\times \text{base}\times \text{height}$
Here we have to find the radius of a circle inscribed in a triangle of sides 9, 12, 15.
Complete step-by-step answer:
The sides of the triangle given in the question are 9, 12, 15.
Let AB = 9, BC = 12, CA = 15.
Now we will check whether the given triangle is a right angled triangle. For this we will use Pythagoras theorem.
$A{{B}^{2}}+B{{C}^{2}}=A{{C}^{2}}$
Substituting the corresponding values, we get
$\begin{align}
& {{9}^{2}}+{{12}^{2}}={{15}^{2}} \\
& \Rightarrow 81+144=225 \\
& \Rightarrow 225=225 \\
\end{align}$
Hence, we can see that the given triangle satisfies Pythagoras theorem, so the given triangle is a right angled triangle. So, the corresponding diagram will be,
So, let the circle inscribed in the triangle ABC have the radius as ‘r’ and ‘O’ be the centre of the circle.
WE can see from the figure that the radius of the inscribed circle is perpendicular to the corresponding sides, so OD, OF, OE are perpendicular to AB, BC and AC respectively.
Now from figure, we can also say that
Area of triangle ABC = Area of triangle AOB + Area of triangle BOC + Area of triangle COA
Now we know the area of the triangle = ½ times base times height. So we can write it as,
$\begin{align}
& \Delta ABC=\Delta AOB+\Delta BOC+\Delta COA \\
& \Rightarrow \dfrac{1}{2}\times AB\times BC=\dfrac{1}{2}\times OD\times AB+\dfrac{1}{2}\times OF\times BC+\dfrac{1}{2}\times OE\times AC \\
\end{align}$
Substituting values from the above figure, we get
$\Rightarrow \dfrac{1}{2}\times 9\times 12=\dfrac{1}{2}\times r\times 9+\dfrac{1}{2}\times r\times 12+\dfrac{1}{2}\times r\times 15$
Cancelling the like terms, we get
$\begin{align}
& \Rightarrow 108=9r+12r+15r \\
& \Rightarrow 108=36r \\
& \Rightarrow r=\dfrac{108}{36}=3 \\
\end{align}$
Hence the radius of the inscribed circle is 3.
Note: Another approach for this problem is using the formula,
$\text{radius}=\left( \dfrac{a+b-c}{2} \right)$
Here a and b are the sides and c is the hypotenuse of the right angled triangle.
This is used when the circle is inscribed in a right angled triangle.
Here we have to find the radius of a circle inscribed in a triangle of sides 9, 12, 15.
Complete step-by-step answer:
The sides of the triangle given in the question are 9, 12, 15.
Let AB = 9, BC = 12, CA = 15.
Now we will check whether the given triangle is a right angled triangle. For this we will use Pythagoras theorem.
$A{{B}^{2}}+B{{C}^{2}}=A{{C}^{2}}$
Substituting the corresponding values, we get
$\begin{align}
& {{9}^{2}}+{{12}^{2}}={{15}^{2}} \\
& \Rightarrow 81+144=225 \\
& \Rightarrow 225=225 \\
\end{align}$
Hence, we can see that the given triangle satisfies Pythagoras theorem, so the given triangle is a right angled triangle. So, the corresponding diagram will be,
So, let the circle inscribed in the triangle ABC have the radius as ‘r’ and ‘O’ be the centre of the circle.
WE can see from the figure that the radius of the inscribed circle is perpendicular to the corresponding sides, so OD, OF, OE are perpendicular to AB, BC and AC respectively.
Now from figure, we can also say that
Area of triangle ABC = Area of triangle AOB + Area of triangle BOC + Area of triangle COA
Now we know the area of the triangle = ½ times base times height. So we can write it as,
$\begin{align}
& \Delta ABC=\Delta AOB+\Delta BOC+\Delta COA \\
& \Rightarrow \dfrac{1}{2}\times AB\times BC=\dfrac{1}{2}\times OD\times AB+\dfrac{1}{2}\times OF\times BC+\dfrac{1}{2}\times OE\times AC \\
\end{align}$
Substituting values from the above figure, we get
$\Rightarrow \dfrac{1}{2}\times 9\times 12=\dfrac{1}{2}\times r\times 9+\dfrac{1}{2}\times r\times 12+\dfrac{1}{2}\times r\times 15$
Cancelling the like terms, we get
$\begin{align}
& \Rightarrow 108=9r+12r+15r \\
& \Rightarrow 108=36r \\
& \Rightarrow r=\dfrac{108}{36}=3 \\
\end{align}$
Hence the radius of the inscribed circle is 3.
Note: Another approach for this problem is using the formula,
$\text{radius}=\left( \dfrac{a+b-c}{2} \right)$
Here a and b are the sides and c is the hypotenuse of the right angled triangle.
This is used when the circle is inscribed in a right angled triangle.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Define absolute refractive index of a medium
Find out what do the algal bloom and redtides sign class 10 biology CBSE
Prove that the function fleft x right xn is continuous class 12 maths CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Select the correct plural noun from the given singular class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
The sum of three consecutive multiples of 11 is 363 class 7 maths CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How many squares are there in a chess board A 1296 class 11 maths CBSE